A) \[\frac{1}{{{\left( 1+\log x \right)}^{3}}}+c\]
B) \[\frac{1}{{{\left( 1+\log x \right)}^{2}}}+c\]
C) \[\frac{x}{\left( 1+\log x \right)}+c\]
D) \[\frac{x}{{{\left( 1+\log x \right)}^{2}}}+c\] Where c is a constant.
Correct Answer: C
Solution :
[c] Let \[I=\int{\frac{\log x}{{{(1+\log x)}^{2}}}dx}\] Put \[\log x=t\Rightarrow \frac{1}{x}dx=dt\] \[I=\int{\frac{{{e}^{t}}t}{{{(1+t)}^{2}}}dt}=\int{\frac{{{e}^{t}}.(t+1-1)}{{{(1+t)}^{2}}}dt}\] \[=\int{\frac{{{e}^{t}}(1+t)}{{{(1+t)}^{2}}}dt-\int{\frac{{{e}^{t}}}{{{(1+t)}^{2}}}dt}}\] \[=\int{\frac{{{e}^{t}}}{1+t}dt-\int{\frac{{{e}^{t}}}{{{(1+t)}^{2}}}dt}}\] \[=\frac{{{e}^{t}}}{1+t}-\int{-{{e}^{t}}\frac{1}{{{(1+t)}^{2}}}dt-\int{\frac{{{e}^{t}}}{{{(1+t)}^{2}}}dt}}\] \[=\frac{{{e}^{t}}}{1+t}+\int{^{{{e}^{t}}}\frac{1}{{{(1+t)}^{2}}}dt-\int{\frac{{{e}^{t}}}{{{(1+t)}^{2}}}dt}}\] \[=\frac{{{e}^{t}}}{1+t}+c=\frac{x}{1+\log x}+c\]
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